A short shunt compound generator has an armature,shunt field and series field resistances of 0.02 ohm, 80 ohm and 0.03 ohm respectively. The generator's induced emf is 510V when the terminal voltage is 500V. How much load current is delivered under this condition. Ja = IL + Ish 90 -VI VI - 500 0-02 0.03 80 V= 505.924 il14 V bad wrient : IL = S05.924 14-s00 0.03 O-D3 IL= 197. 41A

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our professor posted this problem and the solution but i am not sure i can follow. can someone please explain the whole process to me, the formulas she used, & how she came up with the answer i badly need help thank you so much

A short shunt compound generator has an
armature,shunt field and series field resistances
of 0.02 ohm, 80 ohm and 0.03 ohm respectively.
The generator's induced emf is 510V when the
terminal voltage is 500V. How much load current
is delivered under this condition.
Ja = IL + Ish
50 -VI
VI - 500
0-02
D.03
80
V= 505. 924 i14V
boad winent : IL
S05.924 1114-s00
0.03
O-03
IL = 197. 41A
Transcribed Image Text:A short shunt compound generator has an armature,shunt field and series field resistances of 0.02 ohm, 80 ohm and 0.03 ohm respectively. The generator's induced emf is 510V when the terminal voltage is 500V. How much load current is delivered under this condition. Ja = IL + Ish 50 -VI VI - 500 0-02 D.03 80 V= 505. 924 i14V boad winent : IL S05.924 1114-s00 0.03 O-03 IL = 197. 41A
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